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Why are units in abstract algebra called units? Is it just because they generalize the notion of $-1$ and $1$?, and the like? There's often a sense that units don't matter, when talking about things like irreducibility- is the name unit supposed to trivialize them somehow?

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    $\begingroup$ I think that the idea here is that when one measures something in "units" in the usual sense (e.g. meters, seconds, etc.), one is inherently performing a division of some kind to find how many of unit $u$ fit into $x$. So, the "units" of a ring are precisely the elements by which one may divide. $\endgroup$ – Omnomnomnom Oct 5 '15 at 5:54
  • $\begingroup$ I suspect the terminology (like much terminology in ring theory) originates from number theory, where units were things that "behaved like $1$" in factorizations (of integers, or more generally of elements of extensions of the integers) $\endgroup$ – Eric Wofsey Oct 5 '15 at 5:58
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"Unit" from "unity", from Latin unus meaning "one" (see uno, un, etc. in Romance languages). Thus, "1-like thing". In fact, even 1 itself is sometimes referred to as "unity", as in the term roots of unity.

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In the Sciences a unit is a fixed amount of a quantity in question. Every other amount is a multiple of the fixed amount. In the Sciences the zero amount cannot be a unit because no other amount is a multiple of it. In rings, units are those elements with the property that all other elements are multiples of them. Just like in the Sciences.

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  • $\begingroup$ Is this true about abstract algebra too? $\endgroup$ – Arman Malekzadeh Jul 26 '17 at 19:30
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To build upon what user467419 said: In science a unit is a fixed value for which every other amount of something is a multiple of that fixed value.

In rings, units are those elements with the property that all other elements are multiples of them. Specifically, if $u$ is a unit of a ring $R$, then there is a $v\in R$ such that $uv=vu=1$. Thus, for any $x\in R$, one can rewrite $x$ "in terms of the unit $u$" like so: $x=x(vu)=(xv)u$. Thus, $x$ is a multiple of $u$.

This can be understood well in seeing why a ring $F$ is a field (i.e. a commutative ring in which every nonzero element is a unit) if and only if its only ideals are $\{0\}$ and $F$.

Note every element in a field is a unit $u$. It follows that if we rewrite any element $x\in F$ in terms of that unit, we will get $x=ku$ for some $k\in F$. Now consider that any ideal is a kernel of a ring homomorphism $f:F\to R$. $f$'s kernel could be $\{0\}$, in which case $f$ is injective, and the ideal it corresponds to is $\{0\}$.

Otherwise, some element (and thus unit) $u$ maps to zero. Since we can write any $x$ as $x=ku$, that means $f(x)=f(k)f(u)=0.$ So the kernel is all of $F$.

Therefore the only ideals (homomorphism kernels) of a field are $\{0\}$ and $F$, because every element is a unit, so if one element goes to zero, they all must as well, since every element is a multiple of every other.

More generally, since every element in a ring can be written as a multiple of any unit, we can consider any $x$ in a ring $R$ to be $x=ku$, and the behavior of $x$ under any group homomorphism $g$ is determined by the behavior of $k$ and of $u$ under $g$, since $g(x)=g(k)g(u)$. This explains why $u$ is called a "unit" since we can measure behavior of element $x$ in terms of what coefficient $k$ is in front of $u$ when we express $x$ in terms of $u$.

Further (and I do not know if this has to do with how the term originated), a summary for the intuition of the term may be the following theorem:

Theorem (alternate definition of a unit): $u\in R$ is a unit iff for all $x\in R$, there exists a $v\in R$ such that $(xv)u=x$.

The fowards case of this theorem is proven above, while the backward case is provable letting $x=1$.

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